A25.0 ml sample of water at 90 °c has a mass of 24.12 g. calculate the density of water at this temperature.

solution:

Density of water the density of water is greatest at about four degrees C (39.2° F or 277degrees Kelvin) which is a density of 1.000 kg per liter (62.4 pounds per cubic foot). Liquids expand slightly as their temperature is raised, but liquid water is denser than solid water (ice). That is why ice floats: it is less dense than liquid water. That is due to the crystal structure of ice. When water freezes, its volume increases about nine percent. 277 K. Density rho = mass m / Volume V. Water has a density of 1,000 kg/m3 = 1,000 g/L = 1.000 kg/dm3 = 1.000 kg/L = 1.000 g/cm3 = 1.000 g/mL at the temperature of 3.98 degrees Celsius. Temperature in degrees Celsius and the density of water: 1  

At 4 °C, its density is the greatest and is equal to 1000 kilograms per cubic meter (SI unit). At any other temperature, its density is less than 1000 kilograms per cubic meter. The answer "1.0" is wrong. Water is unique in that it is one of the only substances that actually expands and becomes less dense as the temperature increases. Water molecules (H 2 O) have a special shape and arrangement; this is why ice floats above the warmer water on a frozen lake. To convert grams/cm3 or grams/mL (1 cm 3 = 1 mL) into other units, use these conversions: 1 dry ounce = 28.349 grams 1 fluid ounce = 29.574 mL so to go from grams/cm 3 to dry ounces per fluid ounce, multiply by 0.9586.  

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3